
先化简,再求值 (x+1分之1 + x方-1分之x方-2x+1)÷x+1分之x-1, 其中x=2 40
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1/(x+1)+(x²-2x+1)/(x²-1)除以(x-1)/(x+1)
=1/(x+1)+(x-1)²/[(x-1)(x+1)]除以(x-1)/(x+1)
=1/(x+1)+(x-1)/(x+1)除以(x-1)/(x+1)
=[x/(x+1)]×[(x+1)/(x-1)]
=x/(x-1)
当x=2时,原式=2
=1/(x+1)+(x-1)²/[(x-1)(x+1)]除以(x-1)/(x+1)
=1/(x+1)+(x-1)/(x+1)除以(x-1)/(x+1)
=[x/(x+1)]×[(x+1)/(x-1)]
=x/(x-1)
当x=2时,原式=2
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(x+1分之1 + x方-1分之x方-2x+1)÷x+1分之x-1
=[1/(x+1)+(x-1)²/(x-1)(x+1)]÷(x-1)/(x+1)
=[x/(x+1)]×(x+1)/(x-1)
=x/(x-1) 其中x=2
=2/(2-1)
=2
=[1/(x+1)+(x-1)²/(x-1)(x+1)]÷(x-1)/(x+1)
=[x/(x+1)]×(x+1)/(x-1)
=x/(x-1) 其中x=2
=2/(2-1)
=2
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[1/(x+1)+(x²-2x+1)/(x²-1)] / [(x-1)/(x+1)]
= [1/(x+1)+(x-1)²/[(x-1)(x+1)] ]/ [ (x-1)/(x+1)]
=[x/(x+1)]×[(x+1)/(x-1)]
=x/(x-1)
将x=2代入式中得2/(2-1)=2
= [1/(x+1)+(x-1)²/[(x-1)(x+1)] ]/ [ (x-1)/(x+1)]
=[x/(x+1)]×[(x+1)/(x-1)]
=x/(x-1)
将x=2代入式中得2/(2-1)=2
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